Solve the following equation by factoring. \( 6\left(p^{2}-1\right)=5 p \) Rewrite the equation in a completely factored form. \( \square=0 \) (Type your answer in factored form.)

To solve the equation \(6(p^{2}-1)=5p\), first simplify and rewrite it in standard form. This gives: \[ 6p^2 - 6 - 5p = 0 \] Rearranging the terms, we get: \[ 6p^2 - 5p - 6 = 0 \] Next, we can factor this quadratic equation. We look for two numbers that multiply to \(6 \times -6 = -36\) and add up to \(-5\). These numbers are \(-9\) and \(4\). Therefore, we rewrite the equation as: \[ 6p^2 - 9p + 4p - 6 = 0 \] Grouping the terms: \[ 3p(2p - 3) + 2(2p - 3) = 0 \] Factoring by grouping, we have: \[ (3p + 2)(2p - 3) = 0 \] Now, set each factor to zero: 1. \(3p + 2 = 0 \Rightarrow p = -\frac{2}{3}\) 2. \(2p - 3 = 0 \Rightarrow p = \frac{3}{2}\) So the completely factored form is: \[ (3p + 2)(2p - 3) = 0 \] Thus, rewrite the equation as: \[ (3p + 2)(2p - 3) = 0 \]